293. Flip Game只翻转一步的加减号翻转游戏
[抄题]:
You are playing the following Flip Game with your friend: Given a string that contains only these two characters: +
and -
, you and your friend take turns to flip two consecutive "++"
into "--"
. The game ends when a person can no longer make a move and therefore the other person will be the winner.
Write a function to compute all possible states of the string after one valid move.
For example, given s = "++++"
, after one move, it may become one of the following states:
[ "--++", "+--+", "++--" ]
If there is no valid move, return an empty list []
.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
[一句话思路]:
不要用数组来做,字符串就该用函数
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
不要用数组来做,字符串就该用函数
[一刷]:
(i = s.indexOf("++", i+1))) >= 0 表示角标存在性, 加括号不坏菜 特别是赋值的时候
s.substring(i+2)表示切到尾部为止
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
string题一般都用函数
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
substring函数:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
294. Flip Game II 排列组合,用回溯法
[代码风格] :
class Solution { public List<String> generatePossibleNextMoves(String s) { //ini List<String> res = new ArrayList<>(); //cc if (s.length() == 0) { return res; } //for loop for (int i = -1; (i = s.indexOf("++", i + 1)) >= 0;) { res.add(s.substring(0, i) + "--" + s.substring(i + 2)); } //return return res; } }